 | John Dougall - 1810 - 734 pages
...be equal to twice as many right angles as there are triangles, that is, as the figure has sides, but the sum of all the angles of all the triangles is equal to the sum of all the angles of the pol ygor^ together with the angles at the central point G, which by... | |
 | Charles Hutton - Mathematics - 1812 - 620 pages
...three angles of each of these triangles, is equal to two right angles (th. 17) ; therefore the snm «f the angles of all the triangles is equal to twice as many right angles as the figure has sides. But the sum of all the angles about the point *p> which are so many many of the angles... | |
 | Thomas Leybourn - Mathematics - 1814 - 420 pages
...Hence, by adopting the notation in the question, we have But the sum of the angles of any polygon being equal to twice as many right angles as the polygon has sides, less four; the sum of all the angles of the polygon will be equal to an even number of right angles,... | |
 | Charles Hutton - Mathematics - 1816 - 610 pages
...three angles of each of these triangles, is equal to two right angles (th 17*; therefore the sum of the angles of all the triangles is equal to twice as many iig:>t angles as the figure has sides. But the sum of all the angles about the point r, which are so... | |
 | Charles Hutton - Mathematics - 1822 - 616 pages
...three angles of each of these triangles, is equal to two right angles (th. 17) ; therefore the sum of the angles of all the triangles is equal to twice as many right angles as the figure has sides. But the sum of all the angles about the point p, which are so / many of the angles... | |
 | John Radford Young - Euclid's Elements - 1827 - 228 pages
...to say, the sum of the angles of the polygon, together with those about the point within . it, are equal to twice as many right angles as the polygon has sides ; but those angles which are' about the point, amount to four right angles, (Prop. VI. Cor. 2.) deducting... | |
 | Adrien Marie Legendre - Geometry - 1836 - 394 pages
...figure has sides, wanting four right angles. Hence, the interior angles plus four right .. -i angles, is equal to twice as many right angles as the polygon has aides, and consequently, equal to the sum of the interior angles plus the exterior angles. Taking from... | |
 | Nathan Scholfield - 1845 - 894 pages
...of each as there arc sides of the polygon ; hence, the sum of all the interior and exterior angles is equal to twice as many right angles as the polygon has sides. Again, the sum of all the interior angles is equal to two right angles, taken as many times, less two,... | |
 | William Scott - Measurement - 1845 - 288 pages
...end ought to coincide. Also, the sum of all the angles, together with four right angles, ought to be equal to twice as many right angles as the polygon has sides (Eue. i. 32. cor.). To find the angle contained by two straight lines conceived to be drawn from a... | |
 | Anna Cabot Lowell - Geometry - 1846 - 206 pages
...into as many triangles as it has sides. The sum of all the angles of each triangle = 2 RA, therefore the sum of all the angles of all the triangles is equal to 2 RA multiplied by the number of sides in the figure. But the angles (equal to 4 RA) about the point... | |
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... therefore the sum of the angles of all the triangles is equal to twice as many right angles as the figure has sides. But the sum of all the angles... 
